If  $x+y+z=0$  and $\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|=k\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ , then $k=$

$\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|=xa\left(yz{a}^2-x^{2}bc\right)-yb\left(y^{2}ac-xz{b}^2\right)+zc\left(xy{c}^2-ab{z}^2\right)$

$=xyz{a}^3-x^{3}abc-y^{3}abc+xyz{b}^3+xyz{c}^3-abc{z}^3$

$=xyz\left(a^3+b^3+c^3\right)-abc\left(x^3+y^3+z^3\right)$

$=xyz\left(a^3+b^3+c^3\right)-abc\left(3xyz\right)\left[\because from\left(i\right)\right]$

$=xyz\left[a^3+b^3+c^3-3abc\right]$ -----(ii)

$k\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=k\left[abc-a^3-b^3+abc+abc-c^3\right]$

$=k\left[3abc-a^3-b^3-c^3\right]$

$=-k\left[a^3+b^3+c^3-3ab\right]$ -------(iii)

from (ii) & (iii)

$-k=xyz\Rightarrow k=-xyz$